Lacto bacillus has generation time 60 min. at 300 K and 40 min. at 400 K. Determine activation energy in `(kJ)/(mol)` . (`R = 8.3 J K^ (–1) mol^(–1)` ) `[ log_e (2/3) = - 0.4 ]`

Answer

To determine the activation energy (Ea) for the growth of Lactobacillus, we can use the Arrhenius equation in the form of:

\[ \ln \left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

Where:
- \(k_1\) is the generation time at temperature \(T_1\)
- \(k_2\) is the generation time at temperature \(T_2\)
- \(R\) is the universal gas constant (8.314 J K\(^{-1}\) mol\(^{-1}\))
- \(E_a\) is the activation energy
- \(T_1\) and \(T_2\) are the absolute temperatures in Kelvin

### Step 1: Identify the given values
- \(k_1 = 60\) min (at \(T_1 = 300\) K)
- \(k_2 = 40\) min (at \(T_2 = 400\) K)

### Step 2: Convert the generation times to rate constants
Since the generation time is inversely proportional to the rate constant, we have:
- \(k_1 = \frac{1}{60}\) min\(^{-1}\)
- \(k_2 = \frac{1}{40}\) min\(^{-1}\)

### Step 3: Substitute the values into the equation
Using the equation:
\[ \ln \left(\frac{k_2}{k_1}\right) = \ln \left(\frac{\frac{1}{40}}{\frac{1}{60}}\right) = \ln \left(\frac{60}{40}\right) = \ln \left(\frac{3}{2}\right) \]

### Step 4: Use the relationship between natural log and common log
We can convert the natural logarithm to base 10 logarithm using:
\[ \ln x = 2.303 \log_{10} x \]
Thus:
\[ \ln \left(\frac{3}{2}\right) = 2.303 \log_{10} \left(\frac{3}{2}\right) \]
Given that \(\log_{10} \left(\frac{2}{3}\right) = -0.4\), we have:
\[ \log_{10} \left(\frac{3}{2}\right) = -\log_{10} \left(\frac{2}{3}\right) = 0.4 \]

### Step 5: Calculate the temperature difference
Now we calculate:
\[ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{300} - \frac{1}{400} \]
Finding a common denominator (1200):
\[ \frac{4 - 3}{1200} = \frac{1}{1200} \]

### Step 6: Substitute into the Arrhenius equation
Now substituting back into the equation:
\[ \ln \left(\frac{3}{2}\right) = -\frac{E_a}{R} \left(\frac{1}{1200}\right) \]
Substituting the known values:
\[ 2.303 \times 0.4 = -\frac{E_a}{8.314} \left(\frac{1}{1200}\right) \]

### Step 7: Solve for \(E_a\)
Calculating the left side:
\[ 0.9212 = -\frac{E_a}{8.314} \left(\frac{1}{1200}\right) \]
Rearranging gives:
\[ E_a = -0.9212 \times 8.314 \times 1200 \]
Calculating:
\[ E_a = -0.9212 \times 8.314 \times 1200 \approx 9184.4 \text{ J/mol} \]

### Step 8: Convert to kJ/mol
Finally, converting to kJ/mol:
\[ E_a \approx 9.1844 \text{ kJ/mol} \]

### Final Answer
Thus, the activation energy \(E_a\) is approximately:
\[ E_a \approx 9.18 \text{ kJ/mol} \]